Question

20*(3-2x*2)=40√5*(3x*2-2),then x =?

Answer 1

Given: The correct term is 20^((3-2x^(2))) = (40√5)^((3x^2 - 2))

To find: The value of x?

Solution:

• Now we have given the expression as:

                  20^((3-2x^2)) = (40√5)^((3x^2 - 2))

• Now we can rewrite (40√5)^((3x^2 - 2)) term as:

                  (40√5)^((3x^2 - 2)) = (20^3/2)^(3x^2 - 2)

• So comparing it, we get:

                   20^((3-2x^2)) = (20^3/2)^(3x^2 - 2)

• Now we can see that the base is same , 20, so equating the powers, we get:

                  3-2x^2 = 3/2 (3x^2 - 2)

                  3-2x^2 = (9/2)x^2 - 3

                  (-13/2)x^2 = -6

                  x^2 = 12/13

                  x = ±√ 12/13

Answer:

          So the value of x is ±√ 12/13.